Final answer:
The squirrel, launching itself horizontally off a branch with a velocity of 2.8 m/s from a height of 3.4 meters, will land approximately 2.33 meters from the branch, determined using the equations of motion for free fall and horizontal motion.
Step-by-step explanation:
The question is asking how far from the branch a squirrel will land if it launches itself horizontally with a velocity of 2.8 m/s from a height of 3.4 meters. To solve this, we need to use the physics principles of projectile motion and the equations of motion for free fall. Since the squirrel launches horizontally, its initial vertical velocity is zero. Therefore, we only need to consider the time it takes for the squirrel to fall 3.4 meters to the ground and then use this time to calculate the horizontal distance travelled.
First, to find the time (t) it takes to fall 3.4 meters, we use the equation for free fall:
d = 1/2 g t^2
where d is the distance (3.4 m), and g is the acceleration due to gravity (approx. 9.8 m/s^2). Solving for t gives us:
t = √(2d/g) = √(2 * 3.4 m / 9.8 m/s^2) ≈ √(0.693877551) s ≈ 0.833 s
Now that we have the time, we can compute the horizontal distance (x) the squirrel travels using its horizontal velocity (vx):
x t = 2.8 m/s * 0.833 s ≈ 2.33 m
So, the squirrel will land approximately 2.33 meters from the branch, which corresponds to option C.